In the wikipedia article about the pushforward, it is stated that if $f: M\to N$ is smooth, then it induces a bundle map $df: TM \to TN$. It is then claimed that equivalently $f_*=df$ is a bundle map from $TM$ to the pullback bundle $f^* TN$.

Why is this equivalent? The bundles $TN$ and $f^* TN$ are clearly not the same as they are bundles over different spaces. They could be isomorphic, but is still seems strange that this would hold independently of $f$.

Edit: The full quote is

Equivalently (see bundle map), φ∗ = dφ is a bundle map from TM to the pullback bundle φ∗TN over M, which may in turn be viewed as a section of the vector bundle Hom(TM, φ∗TN) over M. The bundle map dφ is also denoted by Tφ and called the tangent map. In this way, T is a functor.

  • $\begingroup$ $TM$ and $f^*TN$ are both bundles over $M$, not bundles over different spaces. $\endgroup$ – user10354138 Jun 11 at 15:07
  • $\begingroup$ @user10354138 You are right. But whis was only a typo. As can be seen from the context of the question, I am interested in $TN$ and $f^*TN$. I corrected the question accordingly $\endgroup$ – Michael Jun 11 at 15:41
  • $\begingroup$ Please, provide a complete quote, afik, the wikipedia article contains no such claim. (It says, however, correctly, that "Then, applying the differential pointwise to $X$ yields the pushforward $\phi_*X$, which is a vector field along $\phi$, i.e., a section of $\phi^*TN$ over $M$." The word "yields" is not the same as "is". $\endgroup$ – Moishe Kohan Jun 11 at 15:55
  • $\begingroup$ @MoisheKohan: I added the quote $\endgroup$ – Michael Jun 12 at 11:52
  • $\begingroup$ OK, then it is just sloppyness in the article; keep in mind that different parts could have been written by different people. The correct statement is "Then, applying the differential pointwise ..." $\endgroup$ – Moishe Kohan Jun 12 at 13:52

Recall the definition of the pull-back bundle (in the context you are interested in): If $f: M\to N$ is a smooth map, $df: TM\to TN$ is the differential of $f$, then (as a topological space) $$ f^*(TN)=\{(x,\xi): x\in M, \xi\in TN, \xi\in T_{f(x)}N\}\subset M\times TN. $$ The bundle structure on $f^*(TN)$ is given by the projection $$ \pi: (x,\xi)\mapsto x. $$ Now, $df$ defines a bundle map $TM\to f^*(TN)$ which I will denote $Df$: $$ Df: (x,\eta)\mapsto (x, df_x(\eta)), x\in M, \eta\in T_xM. $$ I will leave it to you to check that $Df$ is indeed a morphism of vector bundles $$ Df: TM\to f^*(TN). $$ By abuse of notation, one frequently denotes the morphism $Df$ simply $df$ since the"interesting" parts of these morphisms are the same.

Given a vector field $X\in {\mathfrak X}(M)$, the push-forward $f_*(X)$ is the section $Y$ of $f^*(TN)$ which is given by the formula $$ Y_x= Df(X_x). $$ That's all what there is to it. The wikipedia article commits a minor and common abuse of notation by using the notation $df$ for $Df$.

  • $\begingroup$ Thank you, this written very clearly. Just one last thing: If $f$ is a submersion, does this mean $TN \simeq f^* TM$? $\endgroup$ – Michael Jun 14 at 9:33
  • $\begingroup$ @Michael: No, it does not: Ranks could be different. What you need is a local diffeomorphism. $\endgroup$ – Moishe Kohan Jun 14 at 12:13
  • $\begingroup$ Really? What about if $f: X\times Y\to X$ is a projection? $\endgroup$ – Michael Jun 15 at 10:40
  • $\begingroup$ Yes, really. As I said, you need the equality of ranks of $TN$ and $f^*TM$: It is clearly a necessary condition for isomorphism of these bundles. Note that the bundles $f^*TM$ and $TM$ have equal rank. But rank of the tangent bundle equals the dimension of the manifold. Hence, you are asking for $dim(M)=dim(N)$. But if you have a submersion between manifolds of equal dimension, it has to be a local diffeomorphism. $\endgroup$ – Moishe Kohan Jun 15 at 13:29

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